(a) Use the standard results for summation: \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).

Calculate \(\sum_{r=1}^{n} (3r^2 + 3r + 1) = 3 \sum_{r=1}^{n} r^2 + 3 \sum_{r=1}^{n} r + \sum_{r=1}^{n} 1\).

Substitute the standard results: \(3 \left( \frac{n(n+1)(2n+1)}{6} \right) + 3 \left( \frac{n(n+1)}{2} \right) + n\).

Simplify to get \(\frac{1}{2} n(n+1)(2n+1) + \frac{3}{2} n(n+1) + n = n^3 + 3n^2 + 3n\).

(b) Start with \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3 (r+1)^3}\).

Expand \((r+1)^3 - r^3 = r^3 + 3r^2 + 3r + 1 - r^3 = 3r^2 + 3r + 1\).

Thus, \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{3r^2 + 3r + 1}{r^3 (r+1)^3}\).

Use the method of differences: \(\sum_{r=1}^{n} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right) = 1 - \frac{1}{(n+1)^3}\).

(c) As \(n \to \infty\), \(\frac{1}{(n+1)^3} \to 0\), so the sum converges to 1.